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in Stein's book, we have the following definitions: 1)A subspace S of a Hilbert space H is closed if whenever $(f_n) ⊂ S$ converges to some f ∈ H, then f also belongs to S. 2) The set L^2 is complete if every Cauchy sequence $(f_n)$ in $L^2(\mathbb R)$ converges to a function $f \in L^2(\mathbb R).$

By assumption a Hilbert space H is complete. Now what is the difference between a subspace S of H being complete and being closed? Isn't any convergent (in the norm) sequence of functions, a cauchy sequence? So checking for closedness S is equivalent to checking for completness of S, ie whether every cauchy sequence in S conveges in S(in the norm)?

I will take any help I can get as this is really confusing me. Thanks.

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  • $\begingroup$ Yes, a closed subset of a complete metric space is complete, and conversely, if a subset of a complete metric space is complete, then it's closed. $\endgroup$ – quasi Mar 16 '17 at 0:59
  • $\begingroup$ Thanks, so completeness is stronger than closedness? ie if a metric space is complete then it is closed? And so a closed subset of a Hilbert space is Hilbert? $\endgroup$ – user172377 Mar 16 '17 at 1:13
  • $\begingroup$ Basically, a subspace of a hilbert space is complete iff it is closed? $\endgroup$ – user172377 Mar 16 '17 at 1:17
  • $\begingroup$ Yes, because the hilbert space is complete. $\endgroup$ – quasi Mar 16 '17 at 1:21
  • $\begingroup$ Closedness is a relative property, relative to a containing space. Completeness is a property of the space itself. $\endgroup$ – quasi Mar 16 '17 at 1:22
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If you have $S$ a subspace of a Hilbert space $H$, then $S$ being complete is equivalent to $S$ being closed in $H$.

However, completeness is an absolute property, and closedness is a relative property.

For example, let $c_{00}:=\{x \in \mathbb{R}^{\mathbb{N}} \mid x(n) \neq 0 \text{ for finitely many terms}\}$, with inner product $\langle x,y \rangle$ given by $\sum_\limits n x(n)y(n)$.

Consider the sequence $y_n$ in $c_0$ given by $(y_n)(m)=1/m$ for $m \leq n$ and $0$ for $m>n$. It is clear that $y_n$ is Cauchy. However, $y_n$ can't converge to any $x \in c_{00}$, since the distance from $y_n$ to any given $x$ only grows up after $n$ sufficiently large.

Note that the argument was entirely in $c_{00}$. This illustrate that completeness is an intrinsic property.

However, we can see $c_{00}$ as a subspace of the Hilbert space $l^2:=\{x \in \mathbb{R}^{\mathbb{N}} \mid \sum\limits_n (x(n))^2<\infty\}$. Now, my sequence $y_n$ previously defined clearly converges to $x$ given by $x(n)=1/n$. And $x$ is clearly not in $c_{00}$, hence $c_{00}$ is not closed (and therefore not complete by the statement in the yellow bar).

This may seem more straightforward, however it comes at a cost. We must know beforehand some manageable Hilbert space in which our candidate for completeness lives. This is not so easy in general.

Note also that the advantage of completeness relies heavily on the fact that you can assure a sequence will converge by analysing itself, not finding a candidate beforehand. The situation is analogous: the desire for an intrinsic process.


Let's consider extensively a more elementary example, since you seem to know the concept of metric spaces in the comments. Consider the example discussed above: $\big((0,1),d\big)$, where $d(x,y)=|x-y|$.

Let $x_n=1/n$. Note that $d(x_n,x_m)=|1/n-1/m|< 1/\min\{n,m\}$. Therefore, $x_n$ is Cauchy. Indeed, given $\epsilon>0$, take $N$ such that $1/N < \epsilon$. Therefore, if $n,m>N$, then $\min\{n,m\}>N$ and $d(x_n,x_m)=|1/n-1/m|< 1/\min\{n,m\}<1/N < \epsilon$.

However, $x_n$ does not converge. Indeed, for any $x \in (0,1)$, we have that there exists $N$ such that $x> 1/N$, and therefore, for $n>N$, $d(x_n,x)=|x-1/n|=x-1/n>x-1/N$. That is, for $\epsilon:=x-1/N$, there exists no $n$ etc etc.

It follows, by definition, that $\big((0,1),d\big)$ is not complete.

Note that if $(0,1)$ is a subset of another metric space $(X,d')$ for which $d'|_{(0,1) \times (0,1)}=d$, then given a sequence $x_n \in (0,1)$ and an element $x\in (0,1)$,

$x_n$ is a Cauchy sequence in $\big((0,1),d\big)$ if and only if $x_n$ is a Cauchy sequence in $\big(X,d').$

and

$x_n$ converges to $x$ in $\big((0,1),d\big)$ if and only if $x_n$ converges to $x$ in $\big(X,d').$

It doesn't matter who $X$ is, as long as the induced metric is the metric on $X$. This is precisely why we call completeness an "absolute", or "intrinsic" property: it doesn't depend on where we are, as long as it induces the structure we originally have (obviously, otherwise it would be senseless to compare).

Closedness is very sensitive to the metric/topology of the ambient space. $(0,1)$ is closed on $\big((0,1),d\big)$ (as is any metric space as a subset of itself), but $(0,1)$ is not closed on $\big(\mathbb{R},d_{can}\big)$ for example, even though the metric is the induced one.

To be very explicit, when we talk about completeness, the following phrase is meaningful:

The metric space $(X,d)$ is complete.

When talking about closedness, we need the following phrase in order to have an entire meaningful information:

The subset $A \subset X$ is closed in (X,d).

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  • $\begingroup$ I see, so you're advocating it is easier to work with completeness than closedness? Then if we are in a Hilbert space which is complete, why not just throw away closedness idea or definition in the first place and simply work with completeness property? Or the other way around? Jonas explained why closedness might be easier but you say completeness is usually easier to determine? Also could you take a look at my question posed to quasi in the top of the page comments? I would like to hear as much input as possible. $\endgroup$ – user172377 Mar 16 '17 at 1:34
  • $\begingroup$ @Socchi Firstly, I am not advocating it is easier one way or the other. It often depends on context and the informations you have at hand. I will add more info to the answer, hang on. $\endgroup$ – Aloizio Macedo Mar 16 '17 at 1:49
  • $\begingroup$ @Socchi I've made an update. Please see if there are any doubts left. $\endgroup$ – Aloizio Macedo Mar 16 '17 at 2:13
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    $\begingroup$ Thank you so much for the explanations! It is well appreciated. $\endgroup$ – user172377 Mar 16 '17 at 3:38
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Yes, for a subspace $X$ of a complete metric space $Y$, $X$ is complete as a metric space with the restricted metric if and only $X$ is closed as a subset of $Y$. Even if $Y$ is not complete, completeness of $X$ with the restricted metric implies that $X$ is closed in $Y$, because, as you said, convergent sequences are Cauchy. (But if $Y$ is not complete, then closedness of $X$ does not imply completeness, as even taking $X=Y$ shows.)

But when we're working with complete spaces, why make the distinction? (Is that part of your question?) First of all, the equivalence only applies once you're already sitting in a complete space, so for our "big" space $Y$, we need to use the concept of completeness. But then for a subspace $X$ of a complete space $Y$, why not check completeness instead of closedness? I would say that part of the reason is that it's easier to work with and show closedness, and then you get completeness for free when it is convenient. To show closedness you start by assuming a sequence converges and then just have to show the limit is in your subspace. Working with completeness for subspaces in general would add another unnecessary step of pointing out that there is a limit of a Cauchy sequence because of completeness of the big space, then you still have to show the limit is also in the subspace.

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  • $\begingroup$ Thanks that clears up some of my questions, yes I meant why not just work with completeness all the way through if working in a Hilbert space. Can you take a look at my other question that I wrote as a comment to quasi? $\endgroup$ – user172377 Mar 16 '17 at 1:31
  • $\begingroup$ @Socchi: I just responded directly to your comment above, but I also indirectly address that case in the parenthetical at the end of the first paragraph. Every space is closed in itself. However a given metric space may or may not be complete, which as Aloizio Macedo points out is an "intrinsic" property not depending on which space we're contained in, but depending only on our space and its metric. $\endgroup$ – Jonas Meyer Mar 16 '17 at 1:35

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